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From: "Paul.B.Andersen" <relativity@paulba.no>
Newsgroups: sci.physics.relativity
Subject: Re: The problem of relativistic synchronisation
Date: Wed, 4 Sep 2024 21:38:53 +0200
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Den 04.09.2024 02:32, skrev Richard Hachel:
> Le 03/09/2024 à 17:20, "Paul.B.Andersen" a écrit :
>> Den 03.09.2024 12:03, skrev Richard Hachel:
>>>
>>> We have:
>>>
>>> tA(e1) = 0
>>> tA'(e1)= 0
>>> tA(e2) = 0.75 s
>>
>> OK.
> 
>>> And what's missing:
>>> tA'(e2) = 2.25
>>>
>>
>> This is nonsense, and demonstrates that you
>> don't know what an event is.
>> e2 is the event that clock A and clock B' are adjacent
>>    tA'(e2) is meaningless.

> 
> It is obvious that tA'(e2) has a meaning for A'.
> 
> It is the time at which in his frame of reference (A'), the event E2 
> exists.
> 
> Paul, Paul, you can't say it's meaningless. A little more consideration 
> for the other posters, and please a little more practical intelligence: 
> there is indeed a moment, where, for A, the event e2 exists in his frame 
> of reference, and if A' was synchronized at the start, there will be a 
> time, and only one time of e2 that will be written on his watch.
> 
> We can easily, if we correctly master the notion of relativistic 
> simultaneity and the notions of relative chronotropies,
> reveal what this time written on the clock A' thus synchronized during 
> e1 will be: tA'(e1)=0.

e2 is short for "the event that clock A and clock B' are adjacent"

At this event, tA  = (d/v)⋅√(1−v²/c²) = 0.75 s
and clock B will simultaneously in K show tB = 0.75 s

At this event,  tB' = d/v = 1.25 s,
and clock A' will simultaneously in K' show tA' = 1.25 s

> 
> I beg you not to say that it is absurd or meaningless.
> 
> How to proceed?
> 
> We KNOW that the travel time of A in A'B'

At event e1 tA = 0, at event e2 tA = (d/v)⋅√(1−v²/c²) = 0.75 s
so the travel time for A to go from e1 to e2 is = 0.75 s

> will be equal to the distance 
> A'B' in R' divided by the apparent escape velocity of an object moving 
> at v=0.8c.

And what will you escape from?

A  and B  are moving with the speed v = 0.8c in K'  <-
A' and B' are moving with the speed v = 0.8c in K   ->

Nothing is moving with any other speed than v.
There are no 'apparent speeds'.

> This is unavoidable and it is mathematical.
> Let tA'(e2)=tA'(e1)+(A'B'/Vapp')
> Let tA'(e2)= 0 + 3.10^8/(4/9)c
> tA'(2)=2.25 sec

Nonsense.
At e2, tB' = d/v = 1.25 s,
and clock A' will simultaneously in K' show tA' = 1.25 s

Understand this:
A and B are always simultaneously showing the same in K
A' and B' are always simultaneously showing the same in K'

So how can you imagine that at event e2, when tB' = 1.25 s.
then clock A' should simultaneously show 2.25 s ?

> 
> Please note, I am talking about the exact time when A' 
> actually perceives e2 in direct-live, and which represents 
> the real time of things.

Please note, I am talking about the exact time when Paul
actually perceives that Richard snap his fingers in direct-live,
and which represents the real time of things.

> 
> 
> If we want to judge, and count based on synchronization 
> abstract, based on "the transverse speed of light", it is necessary 
> remove a second of anisochrony between A' and B'. Which gives a time 
> supposed (but false) of 1.25 sec.

Well said!

I have always admired your ability to clearly explain things.
Of course we have to remove a second when we are talking
about "the transverse speed of light".
> 
> Please confirm that you have understood and that you validate, which 
> will allow us to go further and explain all the predictive values ​​that 
> I have already given, but without explaining yet.

I have understood that we have to remove a second based on
"the transverse speed of light".
Based on "the longitudinal speed of light" we would have to
add a second, obviously.

-- 
Paul

https://paulba.no/